JAMB Mathematics 2020 Paper

$\frac{1}{3x}$ + $\frac{1}{2}$x = $\frac{2 + 3x}{6x}$ > $\frac{1}{4x}$

= 4(2 + 3x) > 6x = 12x$^{2}$ - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x - 1 > 0

= x < 0, x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x - 1 > 0 = x > 0

x > $\frac{1}{6}$

Combining solutions in cases (1) and (2)

= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$